SIAM J. MATH. ANAL. Vol. 38, No. 4, pp. 1075–1089 c 2006 Society for Industrial and Applied Mathematics A SOLIDIFICATION PHENOMENON IN RANDOM PACKINGS∗ L. BOWEN† , R. LYONS† , C. RADIN‡ , AND P. WINKLER§ Abstract. We prove that uniformly random packings of copies of a certain simply connected figure in the plane exhibit global connectedness at all sufficiently high densities, but not at low densities. Key words. sphere packing, solidification AMS subject classifications. 52C25, 52C17 DOI. 10.1137/050647785 1. Introduction. The densest way to cover a large area with nonoverlapping unit disks is as in Figure 1, in which the disk centers form the vertices of a triangular lattice. Fig. 1. The densest packing of unit disks in the plane. A packing is a collection of congruent copies of a subset with pairwise disjoint interiors. See [5] for a proof that the above packing is indeed the densest possible for unit disks. It is an old unsolved problem to understand whether densest packings of spheres, simplices, or other shapes, in a Euclidean or hyperbolic space of any dimension, exhibit crystallographic symmetry, such as that of Figure 1. This is the spirit, for instance, of Hilbert’s eighteenth problem; see [5, 10] for background. Using physics models of two- and three-dimensional matter as a guide, we are tempted to try to gain insight into densest packings by considering packings at densities below the maximum. (For an example concerning spheres in R3 , see [9].) In ∗ Received by the editors December 16, 2005; accepted for publication (in revised form) July 19, 2006; published electronically MONTH/DAY/YEAR. This research was partially supported by the National Science Foundation under grants DMS-0406017 and DMS-0352999. The authors also thank the Banff International Research Station for support at a workshop where this research was begun. http://www.siam.org/journals/sima/38-4/64778.html † Department of Mathematics, Indiana University, Bloomington, IN 47405 (lpbowen@indiana.edu, rdlyons@indiana.edu). ‡ Department of Mathematics, University of Texas, Austin, TX 78712 (radin@math.utexas.edu). § Department of Mathematics, Dartmouth College, Hanover, NH 03755 (peter.winkler@ dartmouth.edu). 1075 1076 L. BOWEN, R. LYONS, C. RADIN, AND P. WINKLER effect, we are emphasizing not so much that densest packings and low density packings differ by their symmetry, but rather that they differ in some fundamental geometric fashion. Indeed, it is commonly suggested in the physics literature (see, for instance, [1]) that two-dimensional models of matter do not exhibit crystallographic symmetry, and it is sometimes said by mathematicians that in high-dimensional Euclidean space, densest packings of spheres may not have crystallographic symmetry. So perhaps it is appropriate to re-examine the precise manner in which densest packings differ fundamentally from low density packings, and to use packings at less than optimum density as a guide. In this work we replace round disks with deformed disks, which are copies of a “zipper” tile; see Figure 2. This tile can cover the plane completely, in which case the packing has density 1, and is completely connected in any sense. What we show is that even at somewhat lower densities, the uniform random packing still has rigid structure; in particular it has a form of connectedness associated with site percolation [7]. What this means for packing large but finite boxes (with torus boundary conditions) is that the necessary gross irregularities of most packings at such high densities disconnect the packings, if at all, along fault lines whose density tends to 0 as the size of the box tends to infinity. Although we define “uniform random packing” of the plane by limits of measures on packings of finite boxes, the key to our proof is to examine isometryinvariant probability measures on packings of the whole plane and to show that the ones that maximize “degrees of freedom per tile” are unique for high densities. We show that at high density in our model there is a nonzero probability of an infinite linked component, and that this probability is zero at low density. Thus, there are different “phases” of the packings [2]. (This is closely related to continuum percolation, where one looks at overlapping disks with random independent centers, but our methods are quite different.) Although we believe that such a result also holds for packings of disks or of spheres—pairs of which would be called “linked” if sufficiently close—we are able to prove the result only for our tiles, which are shaped to allow three well-defined levels of pairwise separation. (We discuss the appropriate notion of linking for collections of disks in the last section of the paper.) It is generally understood that true crystalline symmetry is not seen below optimal density in two dimensions—see [11]—so the form of connectedness we use may be useful in understanding the role of geometry in Hilbert’s problem. 2. Description of the tile. We consider packings by a deformed disk denoted by t, referred to as “the tile” and depicted in Figure 2. In this section, we define it precisely. Let H be a regular hexagon of area 1. Let r be the radius of the in-circle of H. Let D be a disk concentric with the in-circle and of radius r + ρ, where 0 < ρ 1 is a number we shall choose more precisely later. We shall construct the shape t by modifying H as follows; D will be called the shadow disk of t. As shown in Figure 2, the tile t equals H with each side modified by a “fringe” and each corner modified by a hook and inlet, where a hook is about half an element of the fringe. As shown in Figure 3, the fringe height is 2ρ. The elements of the fringe have two different size “necks,” one of size ρ2 and one of size 2ρ2 , allowing neighboring tiles to be linked in either of two well-defined modes, “tightly linked” and “loosely linked”; the former is illustrated in Figure 4 and the latter in Figure 5. We say that two tiles t are linked (tightly or loosely) if, when one is held fixed, the other can be moved continuously only by a bounded amount (without overlapping the first). A A SOLIDIFICATION PHENOMENON IN RANDOM PACKINGS 1077 Fig. 2. The zipper tile. Fig. 3. Close-up view of the fringe. tight link is one that permits no movement of one tile while fixing the other, while a link that is not tight is called loose. A key feature of our model is that when two tiles are tightly linked, any motion of one would necessitate a corresponding motion of the other. As we shall explain, the uniform probability distribution on packings of the plane at given density is a limit of such a distribution on packings of larger and larger tori. In our model, these distributions on packings of finite tori are concentrated on packings with the largest number of degrees of freedom, and therefore, roughly speaking, the fewest tiles bound by tight links. This gives us useful control on the packings in the support of our distributions. A tile is called fully linked on one side if it is linked with another tile on that side in such a way that either they are tightly linked, and the line joining their centers goes through the midpoint of the sides of the corresponding hexagons, or they are not tightly linked but can be moved continuously so that their shadow disks touch each other. A tile is fully linked if it is fully linked on all sides. We note that the fully tightly linked packing (Figure 4) corresponds to a tiling by the original hexagon and has density 1, and that the tile has area 1 by construction. 1078 L. BOWEN, R. LYONS, C. RADIN, AND P. WINKLER Fig. 4. Tightly linked tiles. 3. Statement of results. To state our results we need some notation. Let X be the space of all packings of the plane by the tile t. Given a compact subset K of the plane and two packings of the plane, we consider the distance between the two packings with respect to K to be the Hausdorff distance between the unions of the tiles in the respective packings intersected with K. Then X is endowed with the topology of Hausdorff convergence on compact subsets; X is compact. Intuitively, two packings are close in X if they are close in the Hausdorff sense in a large ball centered at the origin. We shall define a probability measure on X that is “uniform” on the set of all packings of a fixed density. For this, we shall need the space Xn of all packings by the tile of the n × n torus R2 /(nZ)2 . For any integer m, let Xn,m ⊂ Xn consist of those packings which contain exactly m tiles (Xn,m is empty if m is large enough). To each tile, we assign the set of six unit vectors based on its center and pointing toward the center of each of its edges. Through this assignment, we can view Xn,m as a subset of Tnm /Σm , where Tn is the unit tangent bundle of the n × n torus modulo a 2π/6 rotation and the symmetric group Σm acts by permuting the factors. When m/n2 is small, Xn,m is (3m)-dimensional. However, when m/n2 is sufficiently large, the dimension of Xn,m inside Tnm /Σm is less than 3m. This is because A SOLIDIFICATION PHENOMENON IN RANDOM PACKINGS 1079 Fig. 5. Loosely linked tiles. at least two tiles in any packing of Xn,m will have to be tightly linked, so that it is impossible to move one continuously without moving the other. Thus, it is useful to decompose Xn,m into a (finite) disjoint union of sets Xn,m,k of packings containing exactly k tight links. Generically, the dimension of Xn,m,k is 3(m−k). The dimension can be strictly less than this if the packings are jammed in the sense of [4], although this fact will not be important for us. The top dimension of Xn,m means the maximum dimension of all Xn,m,k . Let μn,m be the probability measure on Xn,m obtained by normalizing the Hausdorff measure on Xn,m in the top dimension of Xn,m with respect to the natural metric inherited from Tnm /Σm . We interpret μn,m as being a uniform measure. The fact that μn,m is supported on those packings with the fewest tight links will be crucial in the analysis to follow. n be the space of all (n × n)-periodic packings of the plane. In other words, Let X n consists of those packings that are preserved under translations by nZ×nZ. Under X the quotient map, this space is naturally identified with Xn . Therefore, we can view n ⊂ X. the measures μn,m as living on X For a fixed density d ∈ [0, 1], let μ(d) be any measure obtained as the weak* limit of measures of the form μn,m such that n → ∞ and m/n2 → d. (Note that m/n2 is the density of every packing in the support of μn,m and d is the average density 1080 L. BOWEN, R. LYONS, C. RADIN, AND P. WINKLER of a packing chosen with respect to μ(d) ; see Lemma 5.1.) A priori, μ(d) may not be unique, although we shall prove that it is for large enough d. A linked component of a packing is a maximal subpacking in which for every two tiles t, t , there is a sequence t = t1 , t2 , . . . , tn = t such that ti is linked to ti+1 (i = 1, . . . , n − 1). A tightly linked component is defined similarly, except that we require ti to be tightly linked to ti+1 . We say that a measure on the space X of packings is invariant if it is preserved under the full isometry group of the plane. All the measures we consider are probability measures unless stated otherwise. Let λ0 be the unique invariant measure on tilings (packings that cover R2 ) by our tile. Let λ1 be the unique invariant measure on packings by t such that all tiles are fully loosely linked, are as close as possible to each other, and the packing has hexagonal symmetry. Write λs := sλ1 + (1 − s)λ0 . Our main results are the following. Theorem 3.1. There exists 0 < d1 < 1 such that if d ≥ d1 , μ(d) is unique and equals λs , where s := (1 − d)/(1 − d1 ). Corollary 3.2. The μ(d) -probability that the origin is inside a tile belonging to an infinite linked component is nonzero for d ≥ d1 . Proposition 3.3. For some d2 > 0, the probability (with respect to any μ(d) for any d < d2 ) that the origin is inside an infinite linked component is zero. 4. Tile properties. Lemma 4.1. For small ρ, if tiles t1 and t2 are not tightly linked and do not overlap, then the distance between their centers is at least 2r + 2ρ. Proof. Consider the line segment from the center of t1 to the center of t2 . If this segment traverses near a corner of t1 or t2 , then it must be longer than 2r + 2ρ for small enough ρ. Suppose it crosses a fringe of t1 and of t2 . If the tiles are not linked, then the claim is obvious. If they are linked, then to minimize the distance, it must be that their fringes match up (so they are fully linked on one side). Thus, they can come closest to each other when pushed flat up against each other so that their shadow disks touch. In this case, the distance between the centers is exactly 2r + ρ. We shall say that two tiles are densely loosely linked if they are loosely linked and their shadow disks touch. There is a unique invariant measure on maximally dense packings by congruent disks [3]. Hence the probability measure λ1 that we defined earlier is the unique invariant measure on packings by t such that all tiles are fully and densely loosely linked. Let d1 be the density of such a packing. Given a tile t in a packing P , we denote by V (t) the Voronoi cell of the center of t with respect to the centers of the other tiles; that is, V (t) is the open set of points closer to the center of t than to the center of any other tile. We denote the area of a region A of the plane by |A|. Lemma 4.2. The following holds for small enough ρ > 0. For any packing P , if t ∈ P is a tile that has no tight links, then the area of V (t) is at least 1/d1 . Moreover, equality holds iff the configuration of tiles determining V (t) is congruent to a corresponding configuration of a packing in the support of λ1 . Proof. For a tile t, let H(t) denote the hexagon from which t is created. For x > 0, let Hx (t) denote the homothetic copy r+x r H(t) about the center of H(t). Suppose t is a tile of P without any tight links. Consider the rays R1 , . . . , R6 from the center of the hexagon H(t) through each of its six vertices. These rays divide the plane into six sectors, S1 , . . . , S6 . A SOLIDIFICATION PHENOMENON IN RANDOM PACKINGS 1081 By construction, if t and t1 are loosely linked, then |Hρ (t) ∩ Hρ (t1 )| = O(ρ2 ): The hexagon interiors do not intersect if they are parallel, while if they are not parallel, they can intersect only very slightly at a corner. The openings at which a corner can enter have area O(ρ2 ) as ρ → 0. Thus, we have proved that whenever t is loosely linked in the sector Si , then |V (t) ∩ Si | ≥ |Hρ (t)|/6 − δ1 , with δ1 = O(ρ2 ) as ρ → 0. Similarly, if t and t2 are not linked at all, then |H2ρ (t) ∩ H2ρ (t2 )| = O(ρ2 ): Again, their interiors do not intersect if they are parallel, while if they are not parallel, they can intersect only very slightly at a corner. So there exists δ2 > 0 such that whenever t is not linked in the sector Si , we have |V (t) ∩ Si | ≥ |H2ρ (t)|/6 − δ2 , with δ2 = O(ρ2 ) as ρ → 0. Therefore, if t has no tight links but is not fully linked, then |V (t)| ≥ j |H (t)| |H (t)| ρ 2ρ − δ1 + (6 − j) − δ2 6 6 for some j with 0 ≤ j ≤ 5. Given that δ1 , δ2 are of order ρ2 while |H2ρ (t)| − |Hρ (t)| is of order ρ, for ρ small enough we may conclude that |V (t)| > |Hρ (t)| in this case. On the other hand, the geometry of a tile is such that for small ρ, if t1 and t2 are two tiles loosely linked to t, then t1 cannot be tightly linked to t2 . Now suppose that t is fully loosely linked. Then the Voronoi cell of the center of t is determined by six tiles t1 , . . . , t6 all loosely linked to t and all with the property that their shadow disks D, D1 , . . . , D6 do not overlap (by the previous lemma). It follows [6] that |V (t)| ≥ |Hρ (t)|, with equality iff each of the disks D1 , . . . , D6 touches D. But there is only one way in which this can occur (up to isometry). So V (t) = Hρ (t) in this case. This implies that the configuration t, t1 , . . . , t6 is congruent to a corresponding configuration of a packing in the support of λ1 . It is easy to see that given ρ > 0, there exists ε > 0 such that for any finite component c of tightly linked tiles in any packing, the union Vc of the Voronoi cells of the centers of the tiles of c has area at least jc + εPerc . Here jc is the number of tiles in c and Perc is the perimeter of the union of hexagons corresponding to c. Let ε be the largest such constant. Let δ > 0 be such that the area of the Voronoi cell in the fully densely loosely linked packing equals 1 + εPer1 + δ, where Per1 is the perimeter of the hexagon of a single tile. Since ε = ρ + O(ρ2 ) and δ = O(ρ2 ), we have the following. Lemma 4.3. For sufficiently small ρ, there are ε, δ > 0 such that for any finite tightly linked component c, 1 , 1 + Per1 ε + δ |Vc | ≥ jc + εPerc , d1 = and δ ≤ ε/100. 5. High density. Recall that X is the compact space of all packings of the plane by the tile (with the topology of Hausdorff-metric convergence on compact subsets). be the space of isometry-invariant Borel probability measures on X. For any Let M , we denote by |μ| := μ(A0 ) the density of μ, where A0 is the set of all packings μ∈M P ∈ X, one of whose tiles contains the origin. Since a tile is the closure of its interior, A0 is a closed set. 1082 L. BOWEN, R. LYONS, C. RADIN, AND P. WINKLER converges to μ in the weak* topology, then |μi | converges Lemma 5.1. If μi ∈ M to |μ|. Proof. Let P denote the union of tiles in a packing, P . For any invariant probability measure ν and any z ∈ R2 , we have |ν| = 1{0∈P } dν(P ) = 1{z∈P } dν(P ). Integrating over z in a unit-area disk, D, with respect to the Lebesgue measure and using Fubini’s theorem gives the identity |ν| = |P ∩ D| dν(P ). Since the function P → |P ∩ D| is continuous on X, the lemma follows. Recall that λ0 is the unique invariant measure on tilings by our tile, so that |λ0 | = 1. Recalling that d1 is the density of a fully densely loosely linked tiling, fix a density d with d1 ≤ d ≤ 1. Let μN be the uniform measure on configuration of tiles at density dN in an N × N torus, where dN → d as N → ∞. To prove Theorem 3.1, we shall show that the weak* limit of μN exists and equals λs , where s := (1−d)/(1−d1 ). We shall use several lemmas that depend on the following notation. Given a packing P ∈ X, let • tP be the tile of P such that the origin belongs to V (tP ) (this exists as long as the origin is not on the boundary of a Voronoi cell), • KP be the tightly linked component containing tP , • jP be the number of tiles in KP , and • f (P ) := 3/jP if jP is finite and tP contains the origin, and 0 otherwise. Thus f (P ), in a sense, measures the number of degrees of freedom per tile near the origin. , then f dν(f ) ≤ 3|ν|, with equality iff Lemma 5.2. If ν is any measure in M tP has no tight links for ν-almost every packing P . The proof is immediate. Lemma 5.3. If a sequence νn ⊂ M converges to ν in the weak* topology, then f dνn converges to f dν. Lemma 5.3 is proved in a manner similar to Lemma 5.1. Given a finite tightly linked component c, let the congruence class of c be C and let XC ⊂ X be the space of all packings P for which tP exists and KP is in C. Let X be the space of all packings P with density 1, where “density” refers to the usual concept of the limit of the proportion of the area of P inside a large disk centered at the origin as the radius tends to infinity. Let X ⊂ X be the space of all packings P such that KP is infinite and either the density of P is less than 1, the density is not defined, or tP does not exist. Thus, X is the disjoint union of X , X and the collection of XC for all C. Let ν be any invariant probability measure with density d. Let νC be ν conditioned on XC , ν be ν conditioned on X , and ν be ν conditioned on X . Since λ0 is the only invariant probability measure with support in X , we have ν = λ0 . Thus, ν = ν(X )ν + ν(X )λ0 + ν(XC )νC . C Define the density |ω| := ω(A0 ) as before, but for any (invariant or noninvariant) probability measure ω on X. We have d = |ν| = ν(X ) |ν | + ν(X ) + ν(XC ) |νC | C A SOLIDIFICATION PHENOMENON IN RANDOM PACKINGS and f dν = ν(XC ) f dνC = C ν(XC ) C 1083 3 |νC |, jC where jC is the number of tiles in C. and C be a finite-component class. Suppose that 0 ≤ s ≤ Lemma 5.4. Let ν ∈ M 1 is such that |λs | = |νC |. Then f dλs ≥ f dνC . Moreover, equality holds only if jC = 1. Proof. As in the proof of Lemma 5.1, wehave that |νC | = jC /|V (tP )| dνC (P ). First suppose that jC =1. Then |νC | = 1/|V (tP )| dνC (P ) ≤ d1 by Lemma 4.2. This means that s = 1 and f dνC = 3 |νC | = 3 |λs | = f dλs . Now assume that j = jC > 1 and put p := PerC . By definition, f dλs = s f dλ1 + (1 − s) f dλ0 = s f dλ1 = 3sd1 . Since νC (f ) = 3|νC |/jC = 3|λs |/jC = 3(sd1 + 1 − s)/j, it suffices to show that sd1 > sd1 + (1 − s) , j which is equivalent to s(jd1 − d1 + 1) > 1. Now sd1 + (1 − s) = |νC | ≤ j j+εp , where ε is from Lemma 4.3. Solving for s gives s≥ 1− j j+εp 1 − d1 , whence it is enough to show that (jd1 − d1 + 1) 1− j j+εp 1 − d1 > 1. This boils down to d1 (pε + 1) > 1. Now, j > 1 implies that p ≥ (7/6)Per1 , where Per1 is the perimeter of a single tile. Since ε/100 > δ (by Lemma 4.3), this implies that pε + 1 > 1 + εPer1 + δ = 1/d1 , proving the last inequality. with |ν| = |λs |. Equality Lemma 5.5. We have f dν ≤ f dλs for all ν ∈ M holds only if • ν(XC ) = 0 for every component class C with jC > 1, and • whenever ν(XC ) > 0 and jC = 1, we have |νC | = d1 . Proof. Recall that f dν = ν(XC ) f dνC . C 1084 L. BOWEN, R. LYONS, C. RADIN, AND P. WINKLER For each component class C, let sC be defined as follows: • If there exists s ∈ [0, 1] such that |νC | = sd1 + (1 − s), then set sC := s; • otherwise, set sC := 1. Let ωC := sC λ1 + (1 − sC )λ0 and ν(XC )ωC . σ := ν(X ) + ν(X ) λ0 + C From the previous lemma, if |νC | ≥ d1 , then f dνC ≤ f dωC , with equality only if jC = 1. If |νC | < d1 , then sC = 1 and 3 |νC | < 3d1 = f dωC . f dνC = jC Summing up, we obtain f dσ = ν(XC ) f dωC C ≥ ν(XC ) f dνC = f dν. C Moreover, equality holds only if ν(XC ) = 0 for every component C with jC > 1 and |νC | = d1 whenever jC = 1. Since |ωC | ≥ |νC |, we have ν(XC ) |ωC | |σ| = ν(X ) + ν(X ) + C ≥ ν(X ) |ν | + ν(X ) + ν(XC ) |νC | C = |ν| = |λs |. of λ0 and λ1 , this implies that f dσ ≤ Since σ and λs are both convex combinations f dλs with equality iff σ = λs . Thus, f dν ≤ f dλs . In the equality case we must have f dν = f dσ = f dλs and σ = λs . This implies that ν(XC ) = 0 if jC > 1 and |νC | = d1 if jC = 1. . If |ν| = |λs |, then f dν ≤ f dλs . Equality holds iff Lemma 5.6. Let ν ∈ M ν = λs . (Informally, λs uniquely maximizes the number of degrees of freedom per tile for invariant measures of a fixed density.) Proof. The previous lemma implies f dν ≤ f dλs . Assume f dν = f dλs ; then ν = ν(X )ν + ν(X )λ0 + ν(XC )νC , where C is the component of size 1 and |νC | = d1 . This gives f dν = ν(XC )3d1 = f dλs = 3sd1 . Hence ν(XC ) = s. Since ν has density strictly less than 1 = |λ0 | but |ν| = |λs |, we must have ν(X ) = 0. That is, ν = ν(X )λ0 + ν(XC )νC . Since ν and λ0 are isometry invariant, νC must also be isometry invariant. By Lemma 4.2, λ1 is the unique isometry-invariant measure with support in XC and with density d1 . Hence νC = λ1 . This implies ν = λs , and the proof is finished. 1085 A SOLIDIFICATION PHENOMENON IN RANDOM PACKINGS Proof of Theorem 3.1. It is easy to see that one can pack the N × N torus in such a way that there is a large region of tightly linked tiles and a large region of densely loosely linked tiles, and in such a way that the interface between the two regions has a density which approaches zero as N tends to infinity, and the density dN of the packing PN tends to d. Let ωN be the invariant measure supported on isometric Then ωN tends to λs in the weak* copies of PN (a pull-back of PN to the plane). topology. By Lemma 5.3, this implies that f dωN → f dλs . Now μN, the uniform measure of density dN on the N × N torus, satisfies f dμN ≥ f dωN . This is because μN is by definition supported on packings with the maximal number of degrees of freedom for the given density dN . Hence lim inf N f dμN ≥ lim inf N f dωN = f dλs . Therefore, if μ∞ is any weak* subsequential limit of μN N , then f dμ∞ ≥ f dλs . But dN → d, so |μ∞ | = |λs | by Lemma 5.1. The previous lemma now implies that μ∞ = λs . Returning to the discussion of the introduction, we note that from simulations of hard disks, one would expect the corollary to hold even for a range of densities below d1 , but we do not know how to prove this. Remark on higher dimensions. The basic features of our argument can be generalized to dimension 3 or higher, except for our use in Lemma 4.2 of [6] on the minimal Voronoi region in disk packings in the plane. It would be of interest if this part of our proof could be replaced with an argument insensitive to dimension, since the Voronoi regions of, say, the spheres in a face-centered cubic lattice do not minimize volume per site [8]. 6. Low density. In this final section, we confirm the intuition that at low densities there will be no infinite loosely linked component. It is obvious that there is no infinite tightly linked component at densities smaller than d1 . We begin with a lemma that holds for any tile shape (in fact, for any collection of shapes and sizes, as long as each can be fit into a disk of some fixed radius s, and “density” is interpreted as the number of tiles per unit area). Lemma 6.1. For small enough density d, if a packing P is drawn from μ(d) , then the probability that the disk BR of radius R about the origin contains more than 9R2 d tile centers goes to zero as R → ∞. Proof. Let s be the radius of the smallest disk containing the tile (in our case, s is about 21/2 · 3−3/4 · (1+2ρ)) and choose 0<d< .05 ; 13πs2 for our zipper tiles with small enough ρ, d ≤ .003 suffices. Let T be the set of tiles whose centers fall in BR , k := πR2 d, and > 9R2 d. Letting μ(d) (·) denote the probability of an event with respect to the measure μ(d) , we shall show that μ(d) (|T | = ) ≤ γ −k μ(d) (|T | = k) for some constant γ < 1. It then follows that μ(d) (|T | > 9R2 d) ≤ μ(d) (|T | = k) ∞ =9R2 d as R → ∞, as desired. γ −k ≤ μ(d) (|T | = k) γ (9−π)R 1−γ 2 d →0 1086 L. BOWEN, R. LYONS, C. RADIN, AND P. WINKLER The measure μ(d) is the limit of uniform distributions of configurations on the N × N torus TN , in turn obtainable by choosing a sequence of n = N 2 d points from the Lebesgue distribution λ on TnN as the centers of the tiles, orienting each tile independently and uniformly at random, and finally conditioning on no overlap. We denote by λ(|T | = j) the a priori probability that exactly j points fall inside BR (which we take to be some fixed disk in the torus). Let Φ be the event that there is no overlap among the tiles whose centers lie in BR , and Ψ be the event that there is no overlap involving any tile whose center falls outside BR . Then λ(|T | = ) λ(Φ |T | = ) λ(Ψ |T | = ∧ Φ) μ(d) (|T | = ) · = · , λ(|T | = k) λ(Φ |T | = k) λ(Ψ |T | = k ∧ Φ) μ(d) (|T | = k) and our job is to bound the three fractions on the right. For the first, we note that |T | is binomially distributed in the measure λ; hence n πR2 N2 λ(|T | = ) = 2 k n πR λ(|T | = k) 1− k < N2 (n−k)−k ( /e)−k n− k (n−k)!/(n− )! n ≤ πR2 n−k !/k! 1− N2 −k k −k ek · = ≤ 0.95−k (n−k)−k 1− πR2 N2 −k k n for large R. The next fraction is easy: Since we may throw the first k centers into BR , then for the remaining −k, we have that λ(Φ |T | = ) λ(Φ |T | = k) is the probability that the additional −k centers do not cause a collision, which is at most 1. For the (inverse of) the third fraction, we throw n − centers into the region outside BR , then throw the remaining − k. A new point, if it lands at a distance greater than 2s from any previous point or from the disk BR , causes no new overlap, and at each stage there are fewer than n−k points already placed. Hence −k 2 λ(Ψ |T | = k ∧ Φ) N − π(R + 2s)2 − (n−k)4πs2 > N 2 − πR2 λ(Ψ |T | = ∧ Φ) ≥ 4πsR + 4πs2 1 − 4πs d − N 2 − πR2 −k 2 −k > 1 − 13s2 d for N R. Putting the inequalities together, we have μ(d) (|T | = ) ≤ μ(d) (|T | = k) .95 1 − 13s2 d where γ := .95/(1 − 13s2 d) < 1 by choice of d. −k = γ −k , A SOLIDIFICATION PHENOMENON IN RANDOM PACKINGS 1087 Proposition 6.2. For some d2 > 0, the μ(d) -probability that the origin is inside an infinite connected component of loosely linked tiles is zero for d < d2 . Proof. Let d ∈ (0, .003) be a density to be chosen later. Let P be a packing drawn from μ(d) ; we aim to show that the probability that the origin is connected by a loosely linked chain of tiles of P to some point at distance R approaches zero as R → ∞. We again choose some large radius R and let T be the set of tiles of P whose centers fall inside the disk BR . Fig. 6. An unlikely configuration of tiles in and around BR . Fix the positions of the tiles of P \ T (the black tiles of Figure 6) and consider the space of packings having these tiles plus n tiles whose centers fall in BR . We think of this space as being a subset of T1 (BR )n /Σn , where T1 (BR ) is the unit tangent bundle of BR (modulo a 2π/6 rotation to take into account the symmetries of the tile) and the symmetric group acts by permuting the factors. If αn is the volume (in T1 (BR )n /Σn -space) of this space and m < n, then by packing n−m tiles into BR and then the remaining m in the leftover space, we have αn ≥ 1 αn−m n n−m m 1 π(R − 2s)2 − nπ(2s)2 , m! where s is, as before, the radius of the circle circumscribing a tile. This takes into account possible intrusion of tiles in P − T into BR , and the fact that a tile center at point x can exclude nearby centers, but only within distance 2s of x. Let β denote the “wiggle room” of a tile t loosely linked to a stationary tile t , that is, the three-dimensional volume of the space of positions of t; then β = O(ρ3 ) (but we use only that β is bounded by a constant). If a packing “percolates,” that is, contains a chain of loosely linked tiles connecting the center to the boundary of 1088 L. BOWEN, R. LYONS, C. RADIN, AND P. WINKLER BR , let t1 , . . . , tm be a shortest such chain (the dark grey tiles of Figure 6). Note that m ≥ R/(2s). For each i > 2, the tile ti is linked to one of the three sides of ti−1 farthest from the side of ti−1 linked to ti−2 , and has wiggle room at most β with respect to ti−1 . Accounting for the orientation of t1 and allowing the remaining n−m tile centers to fall anywhere in BR , we have that the 3n-dimensional volume of the set of percolating packings is bounded by (π/3) · 6 · 3m−2 · β m−1 · αn−m < 3m β m−1 αn−m . Comparing with the lower bound for αn , we find that given |T | = n ≤ 9dR2 , the probability of percolation is less than 3m β m−1 n!/(n−m)! 3m β m−1 αn−m m < ≤ αn π(R − 2s)2 − nπ(2s)2 27βdR2 π (R − 2s)2 − 36dR2 s2 m β, which goes to zero as R (thus also m) increases, for suitably chosen d. Since we know from Lemma 6.1 that μ(d) (|T | ≤ 9dR2 ) approaches 1 as R → ∞, the proposition follows. argument would prove Proposition 6.2 for any density below √ careful A more 1/ 4π(2/3 3) = .2067+ for sufficiently small ρ, but clearly the probability of percolation will remain 0 for much higher densities than that. 7. A conjecture. We have shown that high-density random packings of zipper tiles in the plane contain an infinite loosely linked component with positive probability, while low-density random packings do not. What happens in the case of ordinary disks, where there is no apparent linking mechanism? We believe, but cannot prove, the following. Conjecture. Suppose μ(d) is defined as above, but for geometric disks of radius 1. Join two centers by an edge if their distance is at most 2 + ε for some fixed ε 1. Then for sufficiently high-density d below the maximum, the graph resulting from a configuration drawn from μ(d) will contain an infinite connected component a.s. This connectedness property can in fact be proven by a standard Peierls-type argument for large ε. This may be known already, though we do not know a reference; it is a straightforward extension of the traditional percolation proof to a situation with a new length scale given by the size of the disks. In general, there is some parameter set of (ε, d) ⊂ (0, ∞) × (0, 1) for which there is an infinite component. For small d or for large ε, the problem is quite similar to continuum percolation, where one connects by an edge two points of a Poisson point process if their distance is at most r. Because of homotheties, one may fix the intensity of the point process to be 1. Then there is a phase transition in r. Our situation is quite different in that we really have two parameters, due to the size of the disks, but our conjecture is that there is a phase transition in d for every ε nevertheless. 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